(Last Mod: 07 January 2014 17:32:34 )
For classes in which homework is submitted by hand, the Homework Guidelines are expected to be materially adhered to. Deviations from these guidelines are permitted whenever it is reasonable (as opposed to convenient) to do so. Keep in mind that whether a deviation is reasonable is up to the sole discretion of the grader and/or instructor, both of whom will be biased against doing so -- in other words, you should adhere to the guidelines in all but those rare cases in which deviating is the obviously proper thing to do. Even in the unlikely event that such a deviation is warranted, the goal must be to adhere to the "spirit" of the guidelines.
The key thing to always keep in mind is that your worth to your employer (and the same applies to other venues, as well) will be judged on the quality of your "work product". The concept of "work product" is holistic and encompasses many things, but first and foremost is how well you do your primary job. When it comes to determining raises, promotions, bonuses -- as well as layoffs and firings -- those making the decisions must rely primarily on that subset of your work product that they can actually see or are aware of. Clearly it is in your best interest to make your work product reflect well upon you.
When taking a course, your primary work product are the assignments you submit for grading. Your homework, projects, quizzes, and exams will be the overwhelming determiner of your grade. You should always strive to make what you submit reflect well upon you.
In general, you will get credit for that portion of your work that is demonstrably correct, particularly in terms of the approach taken and less so in terms of the final answer. If you make an error, then an attempt will be made to give credit for those portions that were done correctly but that produce results that are incorrect only because of errors carried forward from previous work. A reasonable effort will be made to only penalize you once for a given error. Thus, it is important to keep in mind that the grader can only grade what you have presented and, in general, will not give you the benefit of the doubt regarding what fraction of the work that you did NOT show was or wasn't correct -- they will grade your work based on how much the submitted work is demonstrably correct.
Remember that the person grading your work is a human being with many demands on his/her time, just like you. If your homework is clear, easy to follow, and well presented then it will be easy to grade. If it contains errors and those errors are easy to spot and determine the scope of, then the grader can award more partial credit. But if the source and scope of the errors if difficult to identify, then the grader is under no obligation to try to decipher your work, or read between the lines, or try to read your mind. The bottom line here is that the quality of your work product is the most telling indicator of how well you value your work and, by extension, your grade for that work. If you don't value your work highly enough to present it well, don't be surprised when the person grading it assigns a commensurate grade.
Make no mistake. You WILL track your units properly or you WILL lose significant credit.
Properly tracking your units in everything that you do will significantly improve your work product. The reason is simple -- we all make mistakes in doing our work and the majority of mistakes will cause problems with the units. If you track your units and regularly inspect them, then you tend to catch these mistakes shortly after they are made and can correct the work quickly and move on. If, at any point, the units do not work out then you KNOW the answer is wrong and any further work you do is just wasted time.
In general the penalties for units-related deficiencies will increase with time. On the first couple of assignments, only minor points will be lost and, mostly, comments will be made. On later assignments the penalties will become progressively harsh. By the middle of the course, it will not be possible to get more than 50% credit for a problem if there are systemic issues with units. An occasional units blunder will not cost you dearly -- it will be treated the same as the occasional math blunder. By the end of the course, if your answer is incorrect because of a mistake that proper tracking of units would have caught, your grade for the problem will start at 25% and go down from there based on the mistakes made. If your answer is wrong because of not tracking units properly, then you will receive no credit for the problem at all.
This sounds draconian largely because it is. But never forget that you are training to be an engineer and engineers are problem solvers. The problems you solve will impact people's lives in ways large and small and in ways that you can't predict at the time you are solving a particular problem.
As an analogy, consider the case of a physician that disregards tracking units when determining the dosage to prescribe for a patient and, as a result, prescribes 2.2 times too much or too little medicine because of not correctly accounting for the need to convert between pounds and kilograms. In one direction they have prescribed a major overdose and in the other they are not prescribing enough medicine to successfully treat the illness. In either case, the patient could easily suffer permanent injury or death. Nearly everyone would agree that such a doctor should be held accountable for the damage caused; few people would have much sympathy for a doctor that couldn't be bothered to do something so simple as track their units to avoid killing a patient. Similarly, few juries would likely have much sympathy in any ensuing trail for malpractice or criminal negligence. In all likelihood, many people would demand that, at the very least, the doctor's license to practice medicine be revoked.
This is a natural reaction because we all know that a careless doctor can kill someone. But now consider that most doctors are intrinsically limited to killing people one at a time. Engineers can and do kill people in job lots. As professionals, we have a moral, ethical, and legal responsibility to exercise all reasonable care to ensure that our work is correct. How much sympathy should an engineer whose work product fails and kills numerous people receive when their error would have been caught had they merely bothered to track their units?
It is astonishing the number of times that someone, be it a student or a practicing professional, will make a mistake of some kind and end up with an answer that is wrong by orders of magnitude and not recognize it because they never bothered to look at their answer and ask if it makes any physical sense. This is another place where engineers have a obligation to exercise reasonable prudence in checking their work. There are four very valuable tools for doing this that all have reasonably low overhead in terms of time and effort; depending on the specifics of the problem, some combination of these should be regularly employed.
First, estimate what the answer should be before you make a detailed effort to solve it. Make whatever simplifications are necessary, either to the problem or to the numbers involved, in order to get a quick answer to a simpler problem that should have an answer similar to the real problem. You should also try to consider how close your estimate probably is to the correct answer. With practice, it is amazing how often you can quickly come up with an answer that you expect will be, say, roughly 10% too high. When you arrive at the real answer, compare it to the estimate. Even if your estimate is only good to an order of magnitude, that is more than good enough to detect that an answer of 10µF is wrong if your estimate was for something in the 200nF range.
Second, which is really an enhancement of the first, is to bound the answer, preferably on both sides, as tight as possible. You can do this by simply performing two estimates as above, but doing one that you know will result in an answer that is too high and one that you know will result in an answer that is too low. If your actual answer does not fall within the bounds, you KNOW something is wrong (and, yes, it could be one of your bounds). Don't go further until the discrepancy is resolved.
It is actually amazing how often you discover that you can quickly get a set of bounds that are tight enough that there is no need to actually solve the problem in detail at all!
Third, perform simple sanity checks on the answer even when it was impractical to do a decent estimate. Something as simple as asking if the sign of the answer is reasonable goes a long way toward catching mistakes.
Fourth, we are fortunate that in most engineering problems, the validity of the answer can be determined from the answer itself by treating the answer as a given and seeing if it is actually consistent with the problem it is presumably solving.
While not required to receive full credit (unless specifically stated), it will be to your benefit, perhaps even in the form of extra credit, if you regularly perform these checks and that is evident in the work you submit (including particularly, on exams)
One common mistake that people make when working a problem is to alter one side of an equation in such a way as to destroy the equality relationship indicated by the presence of the equals sign. For instance, someone might start out with
V = I*R
when looking for an unknown current but given V=4V and R=2Ω and then proceed to extend this as
V = I*R = V/R = 4V/2Ω = 2A
The second equals sign in the above compound equation is wrong. I*R is NOT equal to V/R. If you wish to express your chain of reasoning along these lines, then you should use an arrow to indicate "it therefore follows that", as in
V = I*R → V/R = 4V/2Ω = 2A
The bottom line here is very simple. Throughout your work, any place in which an equals sign appears, the expression on the left of the equals sign must truly be equal to the expression on the right of the equals sign.
How tall am I? Am I 6 tall? Or 183 tall? Or 72 tall? Or 1.83 tall?
You can't tell if any or all of these are right, because they are just numbers; however, someone's height is not a number, it is a "length". Like all measurements, the value of a length has two components -- the numerical part of the value and the units part of the value. Most readers can probably guess what units are associated with each of the above numbers, but that actually makes the point -- engineering is NOT about guessing! It is OUR responsibility to communicate our intentions in clear and unambiguous terms such that the reader is not only not required to guess what those intentions were, but that they aren't even allowed the opportunity to infer them.
A very useful place for learning about units and their proper use is the NIST website on units.
The main thing to keep in mind is that the value of a physical quantity is the product of a numerical value and a unit -- the numerical value by itself is just a number and is NOT the value of a physical quantity.
Units that represent the same physical quantity are considered to be "compatible"; hence feet, centimeters, inches, and meters are all compatible units because they all represent the physical quantity of length. However, they are not identical. To be identical, two units have to not only be compatible, but also have the same scaling factor such that two equal physical quantities having identical units have to have equal numerical values. This is most commonly seen when using derived units, such as volts or ohms. For instance
10 V = 10 J/C = 10 Nm/s = 10 AΩ
The units V (volts), J/C (joules-per-coulomb), Nm/s (newton-meters-per-second) and AΩ (ampere-ohms) are all identical units.
ASIDE: This is probably a good time to mention that the unit associated with a physical quantity is sometimes '1', also known as the "dimensionless unit", and, like radians, even has an official name: the neper, whose symbol is Np. So whether the value of a physical quantity that happens to be dimensionless consists solely of the numerical value (i.e., no unit at all) or whether it simply leaves the dimensionless unit implied, is a matter for semantic debate. For our purposes, we can consider the two views to be equivalent most of the time, but we will assume the latter view (the implied dimensionless unit) when it makes our statements about units more general (i.e., makes it so that we don't have to add qualifiers to deal with the special case of when the quantity is dimensionless).
Going back to our example, it is hopefully obvious that
6 ≠ 183 ≠ 72 ≠ 1.83
But, most people would agree that (with allowance for rounding)
6 ft = 183 cm = 72 in = 1.83 m
This, by itself, should convince you that a physical quantity must be expressed as the numerical value in combination with the corresponding unit in order to have physical meaning.
But people consistently forget this when working with quantities and often make mistakes that could (should) have been caught, sometimes with disastrous consequences such as multibillion dollar space probes smashing into planets instead of going into orbit about them or airliners running out of fuel and having to glide to a landing at a closed airport while a car race is being held on the old runways.
A large part of the blame falls on the educational establishment because it, as a rule, fails to recognize the value of properly tracking units and instilling it as part of educational dogma. Arguably, this is because large fractions of the educational establishment have little to no practical experience in the "real world" and thus, particularly in the mathematical sciences where most of us received our indoctrination into how to do "math", have only ever seen units as an annoyance and something that must occasionally be tacked onto the final result of a bunch of math to add the flavor of being a "practical" word problem. This is reinforced by the fact that nearly all engineering textbooks do not track units properly, but keep in mind that most engineering textbooks are written by people within the educational establishment.
Consider a box of height B sitting on a table of height T and we want to know the height of the top of the box relative to the floor (let's call that H). Our formula is simple:
H = B + T
Now, what if the box is 2 ft tall and the table is 29 in tall? We have
H = 2 ft + 29 in
This is a perfectly valid and accurate expression, although we normally would prefer something that has the standard form of a single numerical value combined with a single unit.
We can get this numerous different ways. Here are three:
H = 2 ft + 29 in
= (2 ft)*[(12 in)/(1 ft)] + 29 in
= 24 in + 29 in
= (24 + 29) in
= 53.0 in
H = 2 ft + 29 in
= 2 ft + (29 in)*[(1 ft)/(12 in)]
= 2 ft + 2.417 ft
= (2 + 2.417) ft
= 4.42 ft
H = 2 ft + 29 in
= (2 ft)*[(1 yd)/(3 ft)]*[(91.44 cm)/(1 yd)] + (29 in)*[(2.54 cm)/(1 in)]
= 60.96 cm + 73.66 cm
= (60.96 + 73.66) cm
= 134.6 cm
In the above expressions, notice the following three things:
The physical value is treated precisely as the product of two entities -- a number and a unit -- and that a unit in the numerator can "cancel" an identical unit in the denominator (within the same term) just like numbers can.
Each quantity within square brackets is identically equal to 1; multiplying a term by such a quantity is the same as multiplying that term by 1 -- i.e., it has no effect on the value of that term.
Like numbers, units can be "factored out" -- e.g., 24 in + 29 in = (24+29) in = 53 in
This last point is particularly useful. In general, we could insist that mathematical operations only be applied to pure numbers, as in the example above, but this quickly gets softened to allowing physical quantities to be added if and only they have identical units. The reason is obvious -- if they have identical units, then we can factor that unit out of both values leaving us with the addition of two pure numbers multiplied by the factored-out unit.
Now let's consider a slightly different question. What is the area, A, of the top of the box if it has a width W and a length L? Our formula for this would be
A = W*L
Now let's say that our width is 18in and our length is 1ft.
A = (18 in)*(1 ft)
= (18)*(in)*(1)*(ft)
= (18)*(1)*(in)*(ft)
= (18*1)*(in*ft)
= 18 in•ft
The above sequence again highlights the notion that we could insist on only performing mathematical operations on pure numbers, but would quickly soften it to the usual rule that when multiplying two physical quantities, the numerical values get multiplied and the units get multiplied with the result being the product of these two separate results.
Although odd, the unit inch-foot is a perfectly valid unit of area since all that is required of a unit of area is to be the product of two units of length. As an example of this in common use, consider the usual unit of volume used for small lakes and reserviors, namely the acre-foot, which is the volume of water necessary to cover an area of one acre to a depth of one foot. Since this has the overall dimension of length cubed, it is a valid measure of volume.
It is not uncommon to see things like:
f[MHz] = 5.03 / √(L[mH]•C[pF])
The idea is that this is telling you that if you enter the value of L in millihenries and the value of C in picofarads, that the result will be in megahertz. Engineering texts and handbooks are replete with formulas like this. In almost all cases, however, these are actually the consequence of the improper treatment of units. The parameters f, L, and C are physical quantities, namely frequency, inductance, and capacitance and, as such, are not pure numbers that you "plug in" to a formula. The variables carry the units and as long as the variables are carried and treated properly, there is no need for such notations (though they can, indeed, be useful as a shortcut when working with commonly scaled values).
The more proper way to express this formula would be
f = 5.03 MHz•√(mH•pF)/√(L•C)
f = 5.03 MHz•√[(mH•pF)/(L•C)]
This preserves all of the shortcut information while being completely general and correct. The L and C can be entered using any correct unit, such as microhenries and microfarads, and the result easily scaled accordingly. The most general form of this equation, though, is simply
ω = 1/√(L•C)
The units of ω is radians/second or, since radians is actually dimensionless, just 1/second or, in the most general case, inverse time, since it could be radians/microsecond or radians/hour.
This brings us to a notational aberration that is so common that we have little option but to accept it at face value and abide by its nuances, namely the difference and relationship between radian frequency, ω, and cyclical frequency, f. We use these two symbols as though they are truly incompatible units when, in fact, they are completely compatible, just as inches and meters are. Consider the symbol for the speed of light, namely c, which can be expressed as either 186,000 miles/sec or 300,000 meters/sec (or whatever other unit of speed) as needed. We don't use c only when we want the speed of light in m/s and use a different symbol, say v, when we want mi/s. So why can't a single symbol for frequency be used with the value being expressed in either radians/sec or cycles/sec or degrees/sec or whatever else as needed? The answer is that it could and, arguably, should. However, the unfortunate reality is that a century or two of people being sloppy with their units has led to the convention that ω is expressed in rad/s and f is expressed in cycles/sec (i.e., Hz).
We then use the conversion relationship
ω = 2πf
This is not dimensionally explicit and would be akin to saying
D = 2.54•H
when we wanted to convert between H in inches and D in centimeters. But remember that H is a variable that carries units of length and not necessarily any particular unit of length. This H could be 18 inches or it could be 1.5 ft or 0.5 yards, all of which are equal and any of which could be substituted in to the above formula. But doing so blindly would leave us with the same unit for D, namely inches, feet, or yards, but would have the wrong numeric value because it would incorrectly multiply the numeric value by 2.54 without changing the unit at all. With the expression as given, we are relying on the reader to accurately guess that H is supposed to be in inches and D is supposed to be in centimeters. But engineering is not about guessing -- that is how people get killed.
Using the sloppy approach described above, we could write this relationship as
D[cm] = 2.54•H[in]
While sloppy, it is at least explicit. In a similar fashion, an explicit form of the frequency relationship would be
ω[rad/s] = 2π•f[Hz]
But the proper way to express the length formula above would be
D = [2.54cm/in]•H
Since the factor in square brackets is identically equal to 1, this emphasizes the fact that D and H are, in fact, equal physical quantities. It is also still a true relationship if H is entered in some unit other than inches, albeit you now have a unit for D that is probably not very useful, such as mile-cm/inch. But it is a straightforward exercise to multiply this by appropriate values of 1 in order to reduce it to whatever the desired unit, say cm, happens to be.
The comparable proper way to express the frequency relationship is
ω = [2π radians/cycle]•f
or
ω = [2π/Hz•s]•f
Since 1 Hz is equal to 1 cycle/s, if f has units of Hz, then this results in the w having units of r/s. As before, the quantity in square brackets is identically equal to 1 (since 2π radians is equal to one complete cycle) thus ω and f are equal physical quantities.
As stated before, this particular aberration is so engrained in the field of engineering that we have little choice but to accept it. Thus, if you use ω = 2πf in your work it will not be marked off, though it is still preferred that when you substitute in numeric values for f that you also properly include the unit of 2π, namely rad/cyc or 1/(Hz-s).
It is not uncommon to work a problem in which some of the quantities are known and some are not. In general, the preferred method for working problems like this is to replace the known quantities with symbolic representations and work with completely symbolic expressions until the end. Sometimes, however, this is not very practical. When an expression contains both symbolic variables and explicit quantities, the symbolic variables carry their units and the explicit quantities must include their units explicitly. It's that simple and straightforward, yet there is a tendency for people to start tacking on the units associated with the symbolic quantities for no good or valid reason and, in doing so, actually result in an expression that is dimensionally incorrect.
For instance, let's say that the time constant, τ, of a circuit is the product of an unknown resistance, R, and a known 10µF capacitance. Let's further stipulate that, for whatever reason, it was decided not to represent the capacitance by the symbol C, and thus we have
τ = R • 10µF
Yet there seems to be this nearly irresistible urge to express this, wrongly, as either
τ = R [Ω] • 10µF (WRONG!)
or
τ = 10•R [µs] (WRONG!)
or some other kluged and incorrect expression. Simply let the symbolic quantities carry their units and the explicit quantities contain both the numeric value and the corresponding unit.
Without getting into too much mathematical rigor, a transcendental function is one that can be defined as an infinite power series. For instance, the sine function can be defined as follows:
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ....
What would happen if the argument, x, had anything other than a dimensionless unit? For example, what if x was 2 seconds? The first term would have units of seconda but the third term would have units of seconds-cubed and, consequently, these two terms could not be added together. The bottom line is that the argument to any transcendental function, such as any of the trigonometric functions, the logarithmic functions, or the exponential functions, have to be dimensionless and it is also true that the value produced by any transcendental function is also dimensionless.
Thus, when you see something like an expression for a time varying voltage given as
v(t) = 10cos(t) (WRONG!)
this has two units errors in it. First, t has units of time and, therefore, the argument to the cosine function, a transcendental function, is not dimensionless. Second, the overall expression is dimensionless since it is the product of 10, a dimensionless pure number, and the result of a transcendental function which is, therefore, also dimensionless.
People will often just gloss over these errors with the reasoning that v(t) is "obviously" in "volts" and t is "obviously" in seconds. There are two problems with this approach. First, it is not at all guaranteed that the expression is in volts (as opposed to millivolts or kilovolts) or that t is in seconds (as opposed to minutes or microseconds). But even more to the point, not being dimensionally correct at this stage completely robs you of the error-detecting potential of tracking units from this point on.
This should be expressed as
v(t) = 10V cos(ωt)
or equivalently
v(t) = 10cos(ωt) V
If the frequency is 1 r/s and you need/want to include that explicitly, then the rules outlined above apply and it would be expressed as
v(t) = 10V cos(1 r/s • t)
There is also nothing wrong with the following equation for v(t)
v(t) = 10V cos(10kHz • t + 45°)
The argument of the cosine function in this equation is completely compatible with being dimensionless. The mistake that many people would make would be to multiply whatever numerical value for t was given by 10,000 and then adding 45 to the result. But doing so would, first off, be adding cycles (which is NOT a dimensionless quantity) to degrees (a different non-dimensionless quantity). Remember, two quantities can only be added together if their units are not only compatible, but identical.
Consider the equation
z = a^b
Let's say that a = 2 gallons and b = 24 inches. What does
z = (2 gallons)^(24 inches)
mean?
How would we express it as a numerical value multiplied by a unit?
Whatever it means, it must also be equal to
z = (16 pints)^(2 feet)
z = (16 pints)^(2 feet)
Remembering that units are treated just like any other factor, these are all of the form
a = xc
and
b = yd
where x and y are the numerical parts while c and d are the units of their respective quantities. Hence we have
z = a^b = (xc)^(yd) = x^(yd) • c^(yd) = (x^y)^d • (c^d)^y
as well as all of the other possible permutations that result from the commutivity of multiplication. A cursory examination should convince you that the units involved simply end up in a byzantine mish-mash unless the units of the exponent, namely d, is 1 (dimensionless). If this is the case, then all of these permutations collapse into the following:
z = x^y • c^y
in which x^y is the numerical value and c^y is the unit
Thus, quantities used as exponents in any expression must be dimensionless.
Note that this is true regardless of what the base is and, therefore, it applies when Euler's number, e, is the base. This shouldn't be surprising since we have already established that the argument of any transcendental function has to be dimensionless and raising e to any value is simply a way of expressing exp(value) which is the exponential function which is transcendental. In fact, raising any constant to an exponent is transcendental since it is trivial to show that
x^y = exp(y ln(x))
It is tempting to overgeneralize this result and assert that, since
c^y = exp(y ln(c))
That the unit of the base must also be dimensionless in order to be able to perform ln(c). The difference here is that when we raise a unit to a power, we never, in fact, combine them and we carry them as a unit raised to a power. It could be argued that the reason why we don't actually combine them is precisely because doing so would imply taking the natural log of a non-dimensionless unit, which can't be done.